Question

Matrix method. ..2x-3y=-1 3x+5y=5​

Answer 1

Answer:

$x=\frac{10}{19}, y=\frac{13}{19}$

Step-by-step explanation:

The given system of equations is

$2x-3y=-1$

$3x+5y=5$

We need to solve these equations by matrix method.

Rewrite the equations in matrix form.

$\begin{bmatrix}2&-3\\ \:3&5\end{bmatrix}\begin{bmatrix}x\\ \:y\end{bmatrix}=\begin{bmatrix}-1\\ \:5\end{bmatrix}$

$A=\begin{bmatrix}2&-3\\ \:3&5\end{bmatrix}$

$X=\begin{bmatrix}x\\ \:y\end{bmatrix}$

$B=\begin{bmatrix}-1\\ \:5\end{bmatrix}$

$AX=B$

$A^{-1}AX=A^{-1}B$

$X=A^{-1}B$                .... (1)

Now first find the inverse of matrix A.

$|A|=2\cdot \:5-\left(-3\right)\cdot \:3$

If $M=\begin{bmatrix}a\:&\:b\:\\ c\:&\:d\:\end{bmatrix}$, then

$adjM=\begin{bmatrix}d\:&\:-b\:\\ -c\:&\:a\:\end{bmatrix}$

$adjA=\begin{bmatrix}5&-\left(-3\right)\\ -3&2\end{bmatrix}$

$A^{-1}=\frac{1}{|A|}adjA$

$A^{-1}=\frac{1}{19}\begin{bmatrix}5&-\left(-3\right)\\ -3&2\end{bmatrix}=\begin{bmatrix}\frac{5}{19}&\frac{3}{19}\\ -\frac{3}{19}&\frac{2}{19}\end{bmatrix}$

Using equation (1), we get

$X=\begin{bmatrix}\frac{5}{19}&\frac{3}{19}\\ -\frac{3}{19}&\frac{2}{19}\end{bmatrix}\begin{bmatrix}-1\\ \:5\end{bmatrix}$

$\begin{bmatrix}x\\ \:y\end{bmatrix}=\begin{bmatrix}\frac{5}{19}\left(-1\right)+\frac{3}{19}\cdot \:5\\ \left(-\frac{3}{19}\right)\left(-1\right)+\frac{2}{19}\cdot \:5\end{bmatrix}$

$\begin{bmatrix}x\\ \:y\end{bmatrix}=\begin{bmatrix}\frac{10}{19}\\ \frac{13}{19}\end{bmatrix}$

On comparing both sides we get

$x=\frac{10}{19}, y=\frac{13}{19}$

Therefore, $x=\frac{10}{19}, y=\frac{13}{19}$.

Answer 2

Answer:

Matrix method. ..2x-3y=-1 3x+5y=5